Math 208 Worksheet 5-16-11
cos n  1

n2
n 1

4n
4n  3n
c)


n
n
5n
n 1 3  4
n 1
x 1 y  2

 z in parametric form.
A)Write x  1  t , y  4  3t , z  1  2t in symmetric form B)
3
2
x 1 y 1 z  3


Find an equation of the plane containing the line
and parallel to the plane
3
2
2
6 x  2 y  4 z  10
x  4 y 1 z  3


Find an equation of the line containing (1,2,3) and parallel to
both in parametric
3
2
1

1) Determine convergence/divergence a)
2)
3)
4)

b)
and symmetric.
5) Find an equation of the plane containing P(1, 2, 1), Q(-1, 2, 0), and R(0,0,1).
6) Determine whether the line
x  1  2t , y  3  t , z  3t
and the plane
x  2 y  3z  2
intersect or not.
If they intersect, find the point of intersection.
(hint: Substitute x, y , z of the line equations into the plane equation and see if there is a solution.)
x  1  2t , y  3  t , z  1  3t and the plane x  y  z  1 intersect or not.
Determine whether the lines x  1  2t , y  3  t , z  3t and x  1  s, y  1  s, z  1  s are parallel,
7) Determine whether the line
8)
intersecting , skew (non-intersecting and non-parallel), or identical.
(hint: first see if they intersect by setting the x and y components equal to each other to solve for s
and t. Now substitute them into z –components to see if they are equal.
They are 4 cases: i) intersecting and the direction vectors are parallel: the lines are identical ii)
intersecting and not the direction vectors are not parallel: the lines are are intersecting iii) nonintersecting and the direction vectors are parallel: the lines are are parallel iv) non-intersecting and
the direction vectors are nonparallel: the lines are skew)
Note: different variables are used for the line 1and line 2. If they use the same variable, then only
the question “do they meet at the same time?” can be answered.
9) Determine whether the lines
x  1  2t , y  6  t , z  t  2
and
x  2  s, y  1  s, z  1  s are parallel,
intersecting, skew or identical.
10) Determine whether the planes are perpendicular, parallel or neither. If neither, find the angle
between the planes. (hint: determine whether their normal vectors are perpendicular, parallel or
neither).
a)
b)
c)
3 x  y  2 z  3 and 6 x  2 y  4 z  10
x  3 y  2 z  4 and 4 x  2 y  z  4
x  y  z  2 and 2 x  y  z  1
1) A) Convergent by comparing since
4n
1
n   3n  4 n
lim
1  cos n 1
 2
n2
n
b) Divergent by the test of divergence since
c) convergent by the comparison test since
3n  4 n 4 n  4 n
4

 2 
n
n
5
5
5
n
2) A)Solve for t and set them equal to each other
x 1 
y  4 z 1

3
2
B) set each expression equal t to get
x  1  3t , y  2  2t , z  t
3) By looking at the symmetric form, we see that (-1,1,-3) is in the plane. By looking at the plane
equation, we see that the plane is normal to <6, -2, 4>. Thus
6( x  1)  2( y  1)  4( z  3)  0
4) Recall that the denominators in the symmetric form make the directional vector.
x  1  3t , y  2  2t , z  3  t
or
x 1 y  2

 z 3
3
2
5) First find the vectors PQ and PR using terminal minus initial. Next take their cross product to find
<a,b,c>. Pick any one of the three points for
( x0 , y0 , z0 ) .
PQ= <-2,0,-1>, PR=<-1,-2,0>
Thus
 a, b, c  2,1,4 , ( x0 , y0 , z0 )  (0,0,1)  2 x  y  4( z  1)  0
6) Substitute x, y , z of the line equations into the plane equation and see if there is a solution.
x  2 y  3z  2  (1  2t )  2(3  t )  3(3t )  2  t  1 . There is a solution, so they intersect
when t = 1, at which x  1  2(1)  3, y  3  (1)  2, z  3(1)  3  (3,2,3)
7) Substitute x, y , z of the line equations into the plane equation and see if there is a solution.
x  y  z  1  (1  2t )  (3  t )  (1  3t )  1  3  0 . There is no solution, so they do not intersect .
1  2t  1  s  2t  s  0
8) Setting x comp and y comp equal to each other.
method to solve for s and t, we get
t  2  z  3(2)  6 .
t  2, s  4 .
Using the second line,
. Use the substitution
3  t 1 s  t  s  2
Using the first line equation to get z,
s  4, z  1  s  z  5 .
Since they produce
different z-values, they do not intersect. Their direction vectors <2, -1, 3> and <-1, 1, 1> are not
parallel. Thus they are skew.
9) Setting x comp and y comp equal to each other.
1  2t  2  s  2t  s  1
6  t 1 s  t  s  5
. Use the elimination
t  2, s  3 . Using the first line equation to get z ,
t  2  z  2  2  4 . Using the second line, s  3, z  1  s  z  4 . Since they produce the
method to solve for s and t, we get
same z-values, they intersect. Their direction vectors <2, -1, 3> and <-1, 1, 1> are not parallel.
Thus they are merely intersecting.
10) A) their normal vectors <3, -1, 2> and <6, -2, 4> are parallel so the planes are parallel.
B)_their normal vectors <1, 3, 2> and <4, -2, 1> are perpendicular since their dot product is 0.
C) their normal vectors <1, -1, 1> and <2,1,1> are neither.
6)
 2 

 18 
  cos 1 